A particle of mass `2 xx 10^-5 kg` moves horizontally between two horizontal plates of a charged parallel plate capacitor between which there is an electric field of `200 NC^-1` acting upward. A magnetic induction of 2.0 T is applied at right angles to the electric field in a direction normal to both `vec B and vec v.` If g is `9.8 ms^-2` and the charge on the particle is `10^-6 C,` then find the velocity of charge particle so that it continues to move horizontally

Options

A. `2 ms^-1`
B. `20 ms^-1`
C. `0.2 ms^-1`
D. `100 ms^-1`

Explanation

To solve the problem, we need to find the velocity of a charged particle moving horizontally in the presence of an electric field and a magnetic field such that it experiences no vertical deflection. Here are the steps to arrive at the solution:

### Step 1: Identify the Forces Acting on the Particle
The particle experiences three forces:
1. **Electric Force (F_e)**: This is due to the electric field acting on the charge.
2. **Gravitational Force (F_g)**: This is due to the weight of the particle.
3. **Magnetic Force (F_m)**: This is due to the motion of the charged particle in the magnetic field.

### Step 2: Calculate the Electric Force
The electric force can be calculated using the formula:
\[ F_e = qE \]
Where:
- \( q = 10^{-6} \, \text{C} \) (charge of the particle)
- \( E = 200 \, \text{N/C} \) (electric field strength)

Substituting the values:
\[ F_e = (10^{-6} \, \text{C}) \times (200 \, \text{N/C}) = 2 \times 10^{-4} \, \text{N} \]

### Step 3: Calculate the Gravitational Force
The gravitational force can be calculated using the formula:
\[ F_g = mg \]
Where:
- \( m = 2 \times 10^{-5} \, \text{kg} \) (mass of the particle)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ F_g = (2 \times 10^{-5} \, \text{kg}) \times (9.8 \, \text{m/s}^2) = 1.96 \times 10^{-4} \, \text{N} \]

### Step 4: Set Up the Equation for No Vertical Deflection
For the particle to continue moving horizontally without deflection, the net vertical force must be zero. Therefore, we set up the equation:
\[ F_e = F_g + F_m \]
Where \( F_m \) is the magnetic force given by:
\[ F_m = qvB \]
Where:
- \( B = 2.0 \, \text{T} \) (magnetic induction)
- \( v \) is the velocity of the particle.

Substituting the forces into the equation:
\[ 2 \times 10^{-4} = 1.96 \times 10^{-4} + (10^{-6})v(2.0) \]

### Step 5: Solve for the Velocity \( v \)
Rearranging the equation gives:
\[ 2 \times 10^{-4} - 1.96 \times 10^{-4} = (10^{-6})v(2.0) \]
\[ 0.04 \times 10^{-4} = (10^{-6})v(2.0) \]
\[ 0.04 \times 10^{-4} = 2 \times 10^{-6}v \]

Dividing both sides by \( 2 \times 10^{-6} \):
\[ v = \frac{0.04 \times 10^{-4}}{2 \times 10^{-6}} \]
\[ v = \frac{0.04}{2} \times 10^{2} \]
\[ v = 0.02 \times 10^{2} \]
\[ v = 2 \, \text{m/s} \]

### Final Answer
The velocity of the charged particle so that it continues to move horizontally is:
\[ \boxed{2 \, \text{m/s}} \]